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I'm struggling with a problem which has been asked a lot of times but always in different versions so I'm still not quote sure about my answer. I have a function l:R→R:x↦l(x) which is the result of different smaller functions namely: $$l(x)=f(x)g(x)-g(x)h(x)$$ and $0<x<1$, with $f(x)>0$;$g(x)>0$ and $h(x)>0$.

Let $f(x)$ and $g(x)$ be convex as well as decreasing over the range of $x$ and let $h(x)$ be concave and also decreasing. Is the following statement true?

  1. The product of $f(x)$ and $g(x)$ is concave and the product of $g(x)$ and $h(x)$ is concave.

  2. But because of $-g(x)h(x)$ the second term of $l(x)$ is convex

  3. We are only interested in positive values for $l(x)$ we can say that $f(x)g(x)>g(x)h(x)$, $l(x)$ is concave for all $l(x)>0$

Thank you.

PAS
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  • Did you mean "Let $f(x)$ and $g(x)$..."? – Ofek Gillon Mar 11 '17 at 13:41
  • oh yes. I will edit the post, thank you – PAS Mar 11 '17 at 13:43
  • We need to investigate the second derivative:

    $$(fg)'' = f'' + 2f'g' + g''$$

    Now, $f,g$ are convex so $f'' ,g'' >0 $. Further more, $f,g$ are decreasing so $f' , g' < 0 \Rightarrow f'g'>0 $, meaning that $(fg)''>0$ and the product is convex.

    About the product of $gh$, the statement feels wrong. I think taking $g(x)=e^{-x}$ and $h$ as some sort of $\arctan$ after a translation will do the job

     – Ofek Gillon Mar 11 '17 at 14:05
  • Thank you for your answer and you are totally right with your remark on $f$ and $g$. To be honest I'm not quite sure about the second derivative of $f''$ so I might be wrong about this. The problem is the following: I know that $l(x)$ definitely concave but the extent depends on a lot of different parameters $(a, b, c, d...)$. The second derivative for $l(x)$ is not an option and I can't really apply any other concept like $l(1-\alpha)x+\alpha y)>(1-\alpha) l(x) +\alpha l(y)$ Are there any options left? – PAS Mar 11 '17 at 14:37
  • the second derivative of f(x)g(x) is wrong. It should be: gf" + 2f'g' + fg". Since f > 0 ang g >0, (fg)' is still > 0 – Hass Sep 14 '17 at 12:22

2 Answers2

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The claim is false in general.

However, Exercise 3.32.(a) on p. 119 in the book quoted by Convexity of the product of two functions in higher dimensions provides the following sufficient condition to obtain your result: $f,g: \mathbf{R} \rightarrow \mathbf{R}$ are convex, both nondecreasing (or non increasing) and positive on an interval.

A condition (stronger than convexity) that directly addresses your issue is log-convexity. The set of log-convex functions is closed under product, sum and positive scaling.

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mlc
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  • What about an "inverse" problem when we know that a product of two functions is convex and that one of the functions in the product is also convex (and positive on some interval in R)? What can we say about the other function in the product? – Confounded Jul 03 '18 at 12:20
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Thank you for your answer. I was checking my formula again for different parameter values and it looks pretty bad for my original claim that $l(x)$ is concave, which is off the table now. In general $l(x)$ can look like this (again for different parameter values): https://ibb.co/eKY16F

The original problem (maybe I should make another thread for this) is the following: There is a value for $x$ that maximizes and minimizes $l(x)$ in the first quadrant (so for non-negative values). The range for $x$ is a compact set with $x[0,x°]$ where $x°$ is the zero of the function $l(x)$ which again depends on different parameter values. Using the theorem of Weierstrass I can say that: (I) That there exists a maximum and a minimum for $l(x)$ with the relevant $x$ values in the compact set $x[0, x°]$.

The Problem is: I can't really say something about the uniqueness of the maximum or the minimum of $l(x)$. I can't use the Karush-Kuhn-Tucker-Theorem because $l(x)$ is not concave. Maybe I should check for quasi-concavity?

PAS
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  • Monotone=quasiconcave. Any chance that $l$ is either concave or monotone? – mlc Mar 12 '17 at 13:58
  • Well for some parameter values $l$ is monotone decreasing (and looks convex) but for other values it is hump-shaped (and concave). I hope this answers your question. Both functions in the link are $l$ for different parameter values. – PAS Mar 12 '17 at 14:06
  • If your $l(x)$ is always concave or monotone (of either kind), the function is quasiconcave. There is a version of the KKT theorem that holds for a quasiconcave objective function. – mlc Mar 12 '17 at 16:42
  • Thank you for your answer. Yes, it's the Arrow-Enthoven-Theorem. But I'm still struggling with the sufficient condition for a quasiconcave objective function, especially the proof for a pretty complex function. But as you were saying "a function is quasiconcave if it is monotone" and although the dotted line is convex (see the link) it is always monotone decreasing. So my function is either hump-shaped or monotone decreasing (depending on the parameter values) and therefore it is quasiconcave? (sry if this is a kinda dump question). – PAS Mar 13 '17 at 13:53
  • After looking at it I think my function is quasiconcave. But I don't think the proof for this is quite difficult for my case. – PAS Mar 13 '17 at 14:18
  • Monotone covers both increasing and decreasing. Perhaps you should open a question about $l(x)$, giving more details and asking if someone can come up with a proof. – mlc Mar 13 '17 at 15:02
  • I will post a new question, can I send you the link when I'm done? – PAS Mar 13 '17 at 15:27
  • Send it in, and I'll give it a look. – mlc Mar 13 '17 at 15:28
  • This is the link for the new post: http://math.stackexchange.com/questions/2186179/quasi-concavity-and-unique-solution – PAS Mar 14 '17 at 12:08